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ryzh [129]
3 years ago
10

A spectrophotometer measures the transmittance or the absorbance, or both, of a particular wavelength of light after it has pass

ed through a liquid sample. Before the transmittance or absorbance of the sample is measured, a cuvette filled only with solvent, called the blank, is placed in the spectrophotometer and measured. Select the reason that, after the blank is measured, the cuvette must be placed in the spectrophotometer in the same orientation each time that it is used. O The spectrophotometer will break if the cuvette position is changed during the experiment. O The transmittance of the liquid must be measured in the same place each time. O The cuvette will only fit into the spectrophotometer in one orientation. O The transmittance of the cuvette must be measured in the same place each time.

Chemistry
1 answer:
lutik1710 [3]3 years ago
5 0

Answer:

The cuvette will only fit into the spectrophotometer in one orientation.

Explanation:

There is an arrow always marked on the cuvette that indicates how to place it inside the spectrophotometer. The sides of the cuvette also have small edges that if light passes through them, spurious light can be detected due to poor positioning. The spurious light does not get through the entire sample, if we want to measure the absorbance. With a cuvette in the indicated position, the procedure is measured and its reference value is taken, and then compared with other cuvettes to be used.

Look the picture!

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1‑Propanol ( P⁰ 1 = 20.9 Torr at 25 ⁰C ) and 2‑propanol ( P⁰ 2 = 45.2 Torr at 25 ⁰C ) form ideal solutions in all proportions. L
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Answer : The mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

Explanation : Given,

Vapor presume of 1‑Propanol (P^o_1) = 20.9 torr

Vapor presume of 2‑Propanol (P^o_2) = 45.2 torr

Mole fraction of 1‑Propanol (x_1) = 0.540

Mole fraction of 2‑Propanol (x_2) = 1-0.540 = 0.46

First we have to calculate the partial pressure of 1‑Propanol and 2‑Propanol.

p_1=x_1\times p^o_1

where,

p_1 = partial vapor pressure of 1‑Propanol

p^o_1 = vapor pressure of pure substance 1‑Propanol

x_1 = mole fraction of 1‑Propanol

p_1=(0.540)\times (20.9torr)=11.3torr

and,

p_2=x_2\times p^o_2

where,

p_2 = partial vapor pressure of 2‑Propanol

p^o_2 = vapor pressure of pure substance 2‑Propanol

x_2 = mole fraction of 2‑Propanol

p_2=(0.46)\times (45.2torr)=20.8torr

Thus, total pressure = 11.3 + 20.8 = 32.1 torr

Now we have to calculate the mole fraction of vapor phase 1‑Propanol and 2‑Propanol.

\text{Mole fraction of 1-Propanol}=\frac{\text{Partial pressure of 1-Propanol}}{\text{Total pressure}}=\frac{11.3}{32.1}=0.352

and,

\text{Mole fraction of 2-Propanol}=\frac{\text{Partial pressure of 2-Propanol}}{\text{Total pressure}}=\frac{20.8}{32.1}=0.648

Thus, the mole fraction of vapor phase 1‑Propanol and 2‑Propanol is, 0.352 and 0.648 respectively.

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